OpenStudy (anonymous):

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 3)(t −1)3, t > 0. For what values of t is the velocity of the particle increasing?

1 year ago
OpenStudy (danjs):

the change in position function s(t) w.r.t. time is the Velocity function , V(t) V(t) = s ' (t)

1 year ago
OpenStudy (anonymous):

Huh?

1 year ago
OpenStudy (danjs):

have you done derivatives,

1 year ago
OpenStudy (anonymous):

yep

1 year ago
OpenStudy (danjs):

take the derivative of the position function, that will be the velocity function... if you take a second derivative, that is then the acceleration function

1 year ago
OpenStudy (danjs):

Velocity will be increasing when you have a net acceleration in the same direction you moving

1 year ago
OpenStudy (anonymous):

So when t>1?

1 year ago
OpenStudy (danjs):

it is like the increasing /decreasing first derivative test from calc 1,

1 year ago
OpenStudy (danjs):

is that a cube on the (t - 1)^3 ?

1 year ago
OpenStudy (anonymous):

Yes.

1 year ago
OpenStudy (danjs):

s(t) = (t + 3)(t −1)^3 v(t) = 4*(t - 1)^2 * (t+2) a(t) = 12*(t - 1)*(t + 1) position, velocity, acceleration functions

1 year ago
OpenStudy (danjs):

They ask when v(t) is increasing in value. you can look at the acceleration function a(t) to decide this, since accel is the rate of change of velocity.

1 year ago
OpenStudy (anonymous):

So the answer isn't t>1?

1 year ago
OpenStudy (anonymous):

My choices are 0 < t < 1 t > 1 t > 0 The velocity is never increasing

1 year ago
OpenStudy (danjs):

oh i dont know yet

1 year ago
OpenStudy (danjs):

yeah looks like + acceleration and increased velocity when t is larger than 1

1 year ago
OpenStudy (danjs):

0 to 1, is negative values

1 year ago
OpenStudy (anonymous):

Thank you. Can you also help me with another question?

1 year ago
OpenStudy (danjs):

SURE BRB

1 year ago
OpenStudy (anonymous):

Okie. A balloon rises at the rate of 8 feet per second from a point on the ground 12 feet from an observer. To 2 decimal places in radians per second, find the rate of change of the angle of elevation when the balloon is 9 feet above the ground. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35). Whenever you are ready. cx

1 year ago
OpenStudy (empty):

What"s your attempt?

1 year ago
OpenStudy (anonymous):

I have non. In all honesty, I'm just trying to finish a pretest as fast as I can with a not so shabby grade, so I can get back to my ACT prep.

1 year ago
OpenStudy (anonymous):

none

1 year ago
OpenStudy (danjs):

yeah this related rate prob..

1 year ago
OpenStudy (empty):

so youre just getting dan to answer your question for you? smh

1 year ago
OpenStudy (empty):

It just looks pretty disrespectful to string him along like this, pretending to be following along while he gives you the answer when he's doing this for free out of the goodness of his own heart and all.

1 year ago
OpenStudy (danjs):

|dw:1454709832743:dw|

1 year ago
OpenStudy (anonymous):

I'm also listening and taking into account what he has to say. You make it sound like I;m just about the answer, I do want to see how it is done because I need it, but at the moment I prefer a fast explained and concise answer for future reference. I apologize that my ACT is a bit more important to me. @Empty

1 year ago
OpenStudy (anonymous):

Okay, why is 8 dh/dt?

1 year ago
OpenStudy (danjs):

rises at 8ft per second, that is the rate of change of height h, dh/dt

1 year ago
OpenStudy (anonymous):

Ohhh, so what would be the next step? The picture helps a lot by the way, cx

1 year ago
OpenStudy (danjs):

you need something to relate the angle and the height, it is a right triangle

1 year ago
OpenStudy (anonymous):

Do I just solve or a when h is 9?

1 year ago
OpenStudy (danjs):

THis is dynamic, the height is changing and the angle is changing at the same time always, they ask for how fast the angle is changing at that instant in time, when h=9...

1 year ago
OpenStudy (danjs):

you need some relationship between the angle and the height then take the derivative with respect to time of everything, so you then have a relationship between the rates of change of height and angle...

1 year ago
OpenStudy (anonymous):

I'm confused.

1 year ago
OpenStudy (danjs):

maybe the tangent will work... tan(a) = h / 12

1 year ago
OpenStudy (danjs):

relationship between the angle and the height

1 year ago
OpenStudy (anonymous):

DO I use 9 for h?

1 year ago
OpenStudy (danjs):

, that will tell you the angle 'a' at that instant... we need to get the rate the angle is changing

1 year ago
OpenStudy (danjs):

take the derivative of both sides of the function with respect to time

1 year ago
OpenStudy (anonymous):

Of the tangent equation?

1 year ago
OpenStudy (danjs):

yes, need to use the chain rule or implicit differentiating derivative of tan(a) will be first derivative with respect to angle a, then da/dt to give you d/dt of the function

1 year ago
OpenStudy (danjs):

tan(a) = h/12 \[\sec^2(a) * \frac{ da }{ dt } = \frac{ 1 }{ 12 }*\frac{ dh }{ dt }\]

1 year ago
OpenStudy (anonymous):

I did it through and got 0.425. Does that sound correct>

1 year ago
OpenStudy (danjs):

i dont know

1 year ago
OpenStudy (danjs):

\[\frac{ da }{ dt }=\frac{ 1 }{ 12 }*\cos^2(a)*\frac{ dh }{ dt }\] so tha tis the rate the angle a changes in terms of the rate h is changing

1 year ago
OpenStudy (danjs):

dh/dt is given at 8 ft/s the angle 'a' needed is found from the instant when h=9 from the tangent tan(a)=9/12 a = 0.6435 that is it, calculate the da/dt

1 year ago
OpenStudy (danjs):

yeah da/dt = 0.43 rad / s when the baloon is at h=9 ft

1 year ago
OpenStudy (danjs):

unless i f'ed up

1 year ago
OpenStudy (anonymous):

YAY!!!

1 year ago
OpenStudy (anonymous):

Do you know anything about Rolles Theorem?

1 year ago
OpenStudy (danjs):

all those probs are similar, need a relationship between variables, and the problem usually has it for you... derivative of both sides, and maybe some substitutions

1 year ago
OpenStudy (danjs):

maybe, is that one of those existence theorems... like intermediate value, mean value

1 year ago
OpenStudy (anonymous):

Yep.

1 year ago
OpenStudy (danjs):

i remember the name, i forget what it says

1 year ago
OpenStudy (anonymous):

Thats okay. Thank you for all the help. Have a great day and life. cx

1 year ago
OpenStudy (danjs):

k goodluck

1 year ago
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